Another p fet won't turn off problem

Other Hardware General Electronics
1

I cannot turn off a p-channel mosfet after turning it on.

I have this circuit on a breadboard:

Here's the mosfet's datasheet:

When I first supplied power the mosfet was off. When I shorted the gate gate to ground (represented as a switch) the fet turned on then stayed on after releasing the short.

All logic tells me the fet should turn off after releasing the short since it has a pull up resister.
What am I missing. Tried a second fet. same problem.

2

Show us your image of the circuit on your breadboard.

Note:
Gate-Source Voltage - Continuous ±8V

You have 12 volts.

3

Here are the photos. In the last photo, upper left hand corner the upper black and red go to the 12v battery. The lower black wire goes to the motor. The red wire connected to the drain goes to the motor.
The motor runs when power is applied and it shouldn't.

mld.com/board1.jpg

mld.com/board2.jpg

mld.com/board3.JPG

4

When I shorted the gate gate to ground (represented as a switch) the fet turned on then stayed on after releasing the short.

It is dead, as explained above:

Note:
Gate-Source Voltage - Continuous ±8V
You have 12 volts.

5

In other words, to elaborate:

This MOSFET can only tolerate up to 8V [plus or minus] on it's Gate [i.e. across the Gate and the Source] Any more than that, and the MOSFET will likely be damaged. In your case, 12V is being applied across the Gate and Source. That's a whopping 4V over the ABSOLUTE MAXIMUM. Thus, it's VERY likely that what is happening is, you're blowing out the Gate, which is causing the MOSFET to get "stuck ON".

Here's where, on the Datasheet, it says this:

AbsMaxVGSS_01.png
AbsMaxVGSS_01.png1152×346 41.5 KB

AbsMaxVGSS_01.png
AbsMaxVGSS_01.png1152×346 41.5 KB

6

Ok, that explains it. Thank you.

Would it work if I apply 5V at the gate to keep it off then short to ground to turn it on?

7

You could:

  1. Put a 5V zener across the 10K
  2. put a 5k in series with your switch.

OR

  1. change the 10K to 4.7k
  2. put a xxxk in series with the switch.

xxx K:

Goal is -5V across from source to Gate.

Assume where you have a 10K is a 4.7k.

Current required is 5/4.7k = 1.006 ma

resistor in series with he switch must then drop 12 - 5 = 7V

R = E/I = 7/ 1.006 ma approx = 7k

So xxxk = 7k ( I think a 6.8k is a standard value that should be OK.

8

Thanks, I'll give it a try.

9

20V MOSFET isn't enough for switching a 12V inductive load without any decoupling capacitors present.

You must have decoupling to control the supply spikes, 100uF or more.

I'd suggest using a 30V or more rated part for extra security with +/-20V or better gate-source abs max rating.